Concave Lenses:
Images

In this section we deal with a very important aspect of
concave lenses, namely how and where they form images. First, a
concave lens is thinner in the center than it is near the edges as
shown in this diagram:

DEFINITION 1:

The focal point of a concave lens is the point where light rays
parallel to the axis seem to diverge from after passing through
the lens. The distance from the lens to this point is called the
focal length of the lens. Because the light rays never
really diverge from this point, this is often referred to as a
virtual focal point.

DEFINITION 2:

Light rays originally headed towards the focal point on the
other side of a concave lens are bent, emerging parallel to the
axis. This is just the reverse of the definition above, but will
be very important in determining how and where images are formed
by a concave lens.

BIG NOTE:

A concave lens has two focal points - one on each side. They
are equal distances from the lens. The lens does not need the same
curvature on both sides for this to be true, and it doesn't depend
on the direction the light takes entering the lens. The combined
curvature determines the focal point.

ON TO IMAGES:

In our diagram, we see an object arrow placed some distance in
front of a concave lens. The two focal points are indicated as f'
since they are virtual focal points. We now consider two of the
many light rays being emitted by the tip of the object arrow.

The light ray which was originally parallel to the axis is bent
and appears to now be coming from the first virtual focal point.
If we are on the right side of the lens, that is how it appears to
us. The dotted line is the apparent path while the solid line is
the real path.

In this diagram we see the light ray that was originally headed
towards the far focal point. It is bent and emerges parallel to
the axis. The next diagram puts those two rays together in the
same diagram

From our diagram we see that the two light rays do not come
together, they do not converge. Therefore there is no place where
we could put a screen and see a real image. Let's imagine that we
are on the right-hand side of this lens. Looking through it
towards the object, the light we see is represented by the solid
lines. In our minds we trace them both back to where they seem to
have originated, giving us this diagram:

The apparent source of the two rays we traced is this point
behind the lens as shown by the hand. In fact, if we were to take
all of the light from the tip of the arrow that went through the
lens, all of the rays would appear to have originated in the same
location. This is shown here, where colors help to sort out real
rays (red) from apparent rays (green).

All the rays from the tip of the arrow appear to be coming from
the original intersection point. Likewise, if we were to keep
track of all the rays from the middle of the object they would
appear to come from a point halfway between the axis and the point
above. And all the rays from the tail of the object arrow would
appear to come from the axis. Therefore we conclude that the image
of our original object appears to be located as shown in this
diagram:

So what are the characteristics of our perceived image?

It is smaller than the original object, or we say it is
reduced.

It is closer to the lens than the original object, but it's
on the same side.

It is not a real image as it cannot be focused onto a
screen.

We call it a virtual image to represent that it is
an apparent image.

The image is right-side up. We call this upright as
opposed to inverted.

GEOMETRY TIME:

It might surprise us to find that the same equation that
governs the images in convex lenses works for concave lenses. But
this is PHYSICS, so nothing should be too surprising. The diagrams
which preceded are re-drawn to show first the quantities we would
typically measure. Then several triangles are drawn which turn out
to be similar. When we are done, we will consider what to do with
positive and negative quantities.

Quantity

Meaning

Do

Distance from lens to object

Ho

Size of object (height)

Di

Distance from lens to image

Hi

Size of image (height)

f

Focal Length

In the two triangles formed above, the larger one being the
lighter shading plus the darker, we see similar triangles if the
object and image are perpendicular to the axis. From the geometry
of similar triangles, we get the following relationship:

Hi / Ho = Di / Do

We recognize this as the magnification equation stating the
ratio of image size to object size. Because we get reduced images,
the magnification will always be less than 1. There is a second
problem, however. The image location is on the same side as the
object, and is determined to be negative. The end result is a
slight modification of our equation:

Hi / Ho = -Di / Do

Now we see similar triangles if we extend the light ray heading
toward the far focal point. The basic relationship will be
Ho/Hi = (Do + f)/f. But the focal length of a concave lens
is negative because it is a virtual focal point rather than a real
one. Therefore our measurement for f must have -f
substituted, becoming:

Ho / Hi = (Do - f) / -f

In these two similar triangles, the larger being the lighter
plus the darker shading, we get Ho/Hi = f/(f-Di). The focal
length of a concave lens is negative because it is a virtual focal
point rather than a real one. Therefore our measurement for
f must have -f substituted. Additionally, the image
is formed on the same side of the lens as the object, so Di is a
negative number, making the measurement we want (-f + Di)
as:

Ho / Hi = -f / (-f +
Di)

We set the two right-hand quantities equal to each other since
they are equal to the same left-hand quantity:

(Do - f) / -f = -f / (-f +
Di)

Cross multiply and expand the binomials:

f2 = (Do - f) (-f + Di) = -f
Do - f Di + f2 + Do Di

Subtract the f2 terms, move the DoDi
term to the left while factoring out -f

-Do Di = -f (Di +
Do)

We now do some mathematical magic. We take the negative of both
sides, divide both sides by f and also by Do Di.
Then we simplify the remaining terms:

1/f = (Di + Do) / DoDi = Di/DoDi +
Do/DoDi = 1/Do + 1/Di

1/f = 1/Do + 1/Di

This equation is exactly the same as the one for convex lenses.
However, when applying it to concave lenses, we note that the
focal length must always be considered negative, and the
arithmetic will generally result in Di being negative also.
Note the next section which discusses signs for lenses forming
images. It is very important.

SIGNS:

For lenses, we develop a sense of signs in our mathematics. It
comes from the direction that light travels through the lens. In
the diagram which follows, note that the positive sense of things
occurs when light starts on one side and converges on the other.

Problem 2

The image formed by a concave lens for an object 40 cm away is
reduced to one fourth its size. What is the focal length of the
lens?

Givens:

Do = 40 cm
Hi = 0.25 Ho

Unknown:

f = ?

Equations:

1/f = 1/Do + 1/Di
Hi / Ho = -Di / Do

Solve:

Hi / Ho = 0.25 Ho / Ho = 0.25 = -Di / Do ..... Di =
-0.25 Do

Di = -0.25 x 40 cm = -10 cm

1/f = 1/Do + 1/Di = 1/40 cm + 1/-10 cm = -0.075 cm-1

f = -13 cm

Problem 3

Jessica looks through a concave lens with a focal length of -20
cm at a tree that's a long distance away. We assume the distance
is very large, or GBN. (Ask your math teacher.) Where is the image
of Jessica's tree formed?

Givens:

Do = GBN or °
f = -20 cm

Unknown

Di = ?

Equation:

1/f = 1/Do + 1/Di

Solve:

1/Di = 1/f - 1/Do = 1/(-20cm) - 1/° = 1/(-20 cm) -
0 = 1/(-20 cm)

Di = -20 cm, or the image
is formed at the focal distance from the lens on the same
side as the tree. Jessica will be looking through the lens
and focusing her eye 20 cm on the other side.